Clustering and classification

A lot of fun, surely.

1. Data

Tasks 1-3.

First, access the necessary libraries.

# Access the needed libraries:
library(dplyr)
## 
## Attaching package: 'dplyr'
## The following objects are masked from 'package:stats':
## 
##     filter, lag
## The following objects are masked from 'package:base':
## 
##     intersect, setdiff, setequal, union
library(tidyr)
library(ggplot2)
library(boot)
library(MASS)
## 
## Attaching package: 'MASS'
## The following object is masked from 'package:dplyr':
## 
##     select
library(tidyverse)
## -- Attaching packages -------------------------------------- tidyverse 1.2.1 --
## <U+221A> tibble  1.3.4     <U+221A> purrr   0.2.4
## <U+221A> readr   1.1.1     <U+221A> stringr 1.2.0
## <U+221A> tibble  1.3.4     <U+221A> forcats 0.2.0
## -- Conflicts ----------------------------------------- tidyverse_conflicts() --
## x dplyr::filter() masks stats::filter()
## x dplyr::lag()    masks stats::lag()
## x MASS::select()  masks dplyr::select()
library(corrplot)
## corrplot 0.84 loaded

Let’s load the Boston data from the MASS package and explore the structure and the dimensions of the data and describe the dataset.

# load the data
data("Boston")

# explore the dataset
str(Boston)
## 'data.frame':    506 obs. of  14 variables:
##  $ crim   : num  0.00632 0.02731 0.02729 0.03237 0.06905 ...
##  $ zn     : num  18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
##  $ indus  : num  2.31 7.07 7.07 2.18 2.18 2.18 7.87 7.87 7.87 7.87 ...
##  $ chas   : int  0 0 0 0 0 0 0 0 0 0 ...
##  $ nox    : num  0.538 0.469 0.469 0.458 0.458 0.458 0.524 0.524 0.524 0.524 ...
##  $ rm     : num  6.58 6.42 7.18 7 7.15 ...
##  $ age    : num  65.2 78.9 61.1 45.8 54.2 58.7 66.6 96.1 100 85.9 ...
##  $ dis    : num  4.09 4.97 4.97 6.06 6.06 ...
##  $ rad    : int  1 2 2 3 3 3 5 5 5 5 ...
##  $ tax    : num  296 242 242 222 222 222 311 311 311 311 ...
##  $ ptratio: num  15.3 17.8 17.8 18.7 18.7 18.7 15.2 15.2 15.2 15.2 ...
##  $ black  : num  397 397 393 395 397 ...
##  $ lstat  : num  4.98 9.14 4.03 2.94 5.33 ...
##  $ medv   : num  24 21.6 34.7 33.4 36.2 28.7 22.9 27.1 16.5 18.9 ...
summary(Boston)
##       crim                zn             indus            chas        
##  Min.   : 0.00632   Min.   :  0.00   Min.   : 0.46   Min.   :0.00000  
##  1st Qu.: 0.08204   1st Qu.:  0.00   1st Qu.: 5.19   1st Qu.:0.00000  
##  Median : 0.25651   Median :  0.00   Median : 9.69   Median :0.00000  
##  Mean   : 3.61352   Mean   : 11.36   Mean   :11.14   Mean   :0.06917  
##  3rd Qu.: 3.67708   3rd Qu.: 12.50   3rd Qu.:18.10   3rd Qu.:0.00000  
##  Max.   :88.97620   Max.   :100.00   Max.   :27.74   Max.   :1.00000  
##       nox               rm             age              dis        
##  Min.   :0.3850   Min.   :3.561   Min.   :  2.90   Min.   : 1.130  
##  1st Qu.:0.4490   1st Qu.:5.886   1st Qu.: 45.02   1st Qu.: 2.100  
##  Median :0.5380   Median :6.208   Median : 77.50   Median : 3.207  
##  Mean   :0.5547   Mean   :6.285   Mean   : 68.57   Mean   : 3.795  
##  3rd Qu.:0.6240   3rd Qu.:6.623   3rd Qu.: 94.08   3rd Qu.: 5.188  
##  Max.   :0.8710   Max.   :8.780   Max.   :100.00   Max.   :12.127  
##       rad              tax           ptratio          black       
##  Min.   : 1.000   Min.   :187.0   Min.   :12.60   Min.   :  0.32  
##  1st Qu.: 4.000   1st Qu.:279.0   1st Qu.:17.40   1st Qu.:375.38  
##  Median : 5.000   Median :330.0   Median :19.05   Median :391.44  
##  Mean   : 9.549   Mean   :408.2   Mean   :18.46   Mean   :356.67  
##  3rd Qu.:24.000   3rd Qu.:666.0   3rd Qu.:20.20   3rd Qu.:396.23  
##  Max.   :24.000   Max.   :711.0   Max.   :22.00   Max.   :396.90  
##      lstat            medv      
##  Min.   : 1.73   Min.   : 5.00  
##  1st Qu.: 6.95   1st Qu.:17.02  
##  Median :11.36   Median :21.20  
##  Mean   :12.65   Mean   :22.53  
##  3rd Qu.:16.95   3rd Qu.:25.00  
##  Max.   :37.97   Max.   :50.00

The Boston data frame has 506 rows and 14 columns. It describes housing values in the suburbs of Boston.

What are the variables in the data?

colnames(Boston)
##  [1] "crim"    "zn"      "indus"   "chas"    "nox"     "rm"      "age"    
##  [8] "dis"     "rad"     "tax"     "ptratio" "black"   "lstat"   "medv"

The descriptions of the variables are available here. They concern such things as per capita crime rate by town, average number of rooms per dwelling, or even pupil-teacher ratio by town.

Now let’s have a look at a graphical overview of the data and show summaries of the variables in the data.

summary(Boston)
##       crim                zn             indus            chas        
##  Min.   : 0.00632   Min.   :  0.00   Min.   : 0.46   Min.   :0.00000  
##  1st Qu.: 0.08204   1st Qu.:  0.00   1st Qu.: 5.19   1st Qu.:0.00000  
##  Median : 0.25651   Median :  0.00   Median : 9.69   Median :0.00000  
##  Mean   : 3.61352   Mean   : 11.36   Mean   :11.14   Mean   :0.06917  
##  3rd Qu.: 3.67708   3rd Qu.: 12.50   3rd Qu.:18.10   3rd Qu.:0.00000  
##  Max.   :88.97620   Max.   :100.00   Max.   :27.74   Max.   :1.00000  
##       nox               rm             age              dis        
##  Min.   :0.3850   Min.   :3.561   Min.   :  2.90   Min.   : 1.130  
##  1st Qu.:0.4490   1st Qu.:5.886   1st Qu.: 45.02   1st Qu.: 2.100  
##  Median :0.5380   Median :6.208   Median : 77.50   Median : 3.207  
##  Mean   :0.5547   Mean   :6.285   Mean   : 68.57   Mean   : 3.795  
##  3rd Qu.:0.6240   3rd Qu.:6.623   3rd Qu.: 94.08   3rd Qu.: 5.188  
##  Max.   :0.8710   Max.   :8.780   Max.   :100.00   Max.   :12.127  
##       rad              tax           ptratio          black       
##  Min.   : 1.000   Min.   :187.0   Min.   :12.60   Min.   :  0.32  
##  1st Qu.: 4.000   1st Qu.:279.0   1st Qu.:17.40   1st Qu.:375.38  
##  Median : 5.000   Median :330.0   Median :19.05   Median :391.44  
##  Mean   : 9.549   Mean   :408.2   Mean   :18.46   Mean   :356.67  
##  3rd Qu.:24.000   3rd Qu.:666.0   3rd Qu.:20.20   3rd Qu.:396.23  
##  Max.   :24.000   Max.   :711.0   Max.   :22.00   Max.   :396.90  
##      lstat            medv      
##  Min.   : 1.73   Min.   : 5.00  
##  1st Qu.: 6.95   1st Qu.:17.02  
##  Median :11.36   Median :21.20  
##  Mean   :12.65   Mean   :22.53  
##  3rd Qu.:16.95   3rd Qu.:25.00  
##  Max.   :37.97   Max.   :50.00

From the summary of the variables we can see minimum, maximum, median and mean values as well as the 1st and 3rd quartiles of the variables.

The correlations between the different variables can be studied with the help of a correlations matrix and a correlations plot.

# First calculate the correlation matrix and round it so that it includes only two digits:
cor_matrix<-cor(Boston) %>% round(digits = 2)

# Print the correlation matrix:
cor_matrix
##          crim    zn indus  chas   nox    rm   age   dis   rad   tax
## crim     1.00 -0.20  0.41 -0.06  0.42 -0.22  0.35 -0.38  0.63  0.58
## zn      -0.20  1.00 -0.53 -0.04 -0.52  0.31 -0.57  0.66 -0.31 -0.31
## indus    0.41 -0.53  1.00  0.06  0.76 -0.39  0.64 -0.71  0.60  0.72
## chas    -0.06 -0.04  0.06  1.00  0.09  0.09  0.09 -0.10 -0.01 -0.04
## nox      0.42 -0.52  0.76  0.09  1.00 -0.30  0.73 -0.77  0.61  0.67
## rm      -0.22  0.31 -0.39  0.09 -0.30  1.00 -0.24  0.21 -0.21 -0.29
## age      0.35 -0.57  0.64  0.09  0.73 -0.24  1.00 -0.75  0.46  0.51
## dis     -0.38  0.66 -0.71 -0.10 -0.77  0.21 -0.75  1.00 -0.49 -0.53
## rad      0.63 -0.31  0.60 -0.01  0.61 -0.21  0.46 -0.49  1.00  0.91
## tax      0.58 -0.31  0.72 -0.04  0.67 -0.29  0.51 -0.53  0.91  1.00
## ptratio  0.29 -0.39  0.38 -0.12  0.19 -0.36  0.26 -0.23  0.46  0.46
## black   -0.39  0.18 -0.36  0.05 -0.38  0.13 -0.27  0.29 -0.44 -0.44
## lstat    0.46 -0.41  0.60 -0.05  0.59 -0.61  0.60 -0.50  0.49  0.54
## medv    -0.39  0.36 -0.48  0.18 -0.43  0.70 -0.38  0.25 -0.38 -0.47
##         ptratio black lstat  medv
## crim       0.29 -0.39  0.46 -0.39
## zn        -0.39  0.18 -0.41  0.36
## indus      0.38 -0.36  0.60 -0.48
## chas      -0.12  0.05 -0.05  0.18
## nox        0.19 -0.38  0.59 -0.43
## rm        -0.36  0.13 -0.61  0.70
## age        0.26 -0.27  0.60 -0.38
## dis       -0.23  0.29 -0.50  0.25
## rad        0.46 -0.44  0.49 -0.38
## tax        0.46 -0.44  0.54 -0.47
## ptratio    1.00 -0.18  0.37 -0.51
## black     -0.18  1.00 -0.37  0.33
## lstat      0.37 -0.37  1.00 -0.74
## medv      -0.51  0.33 -0.74  1.00
# Visualize the correlation matrix with a correlations plot:
corrplot(cor_matrix, method="circle", type = "upper", cl.pos = "b", tl.pos = "d", tl.cex = 0.6)

From the plot above we can easily see which variables correlate with which and is that correlation positive (blue) or negative (red). Some observations:

  • accessibility to radial highways (rad) is highly positively correlated with property taxes (tax), as are industrial land use (indus) and air pollution (nox) - land next to the roads is taxed higher because of higher profit posiibilities, there is more industrial plots and obviously air pollution levels are up because of the cars;
  • the distance from the employment centres (dis) correlates negatively with age of the houses (age), nitrogen oxides concentration (nox) and proportion of non-retail business acres (indus) - there are newer houses farther away from the centrally-located workplaces, there is less air pollution in the suburbs, and less land is devoted to businesses other than local retail shops;
  • Charles river dummy variable (chas) does not correlate with any other variable, as is to be expected.

2. Standardization and scaling of the data

Task 4

In this part, we are performing the following: * Standardize the dataset and print out summaries of the scaled data. * Create a categorical variable of the crime rate in the Boston dataset (from the scaled crime rate). * Use the quantiles as the break points in the categorical variable. * Drop the old crime rate variable from the dataset. * Divide the dataset to train and test sets, so that 80% of the data belongs to the train set.

Let’s standardize the dataset and print out summaries of the scaled data for the later classification and clustering analysis. How did the variables change?

# center and standardize variables
boston_scaled <- scale(Boston)

# summaries of the scaled variables
summary(boston_scaled)
##       crim                 zn               indus        
##  Min.   :-0.419367   Min.   :-0.48724   Min.   :-1.5563  
##  1st Qu.:-0.410563   1st Qu.:-0.48724   1st Qu.:-0.8668  
##  Median :-0.390280   Median :-0.48724   Median :-0.2109  
##  Mean   : 0.000000   Mean   : 0.00000   Mean   : 0.0000  
##  3rd Qu.: 0.007389   3rd Qu.: 0.04872   3rd Qu.: 1.0150  
##  Max.   : 9.924110   Max.   : 3.80047   Max.   : 2.4202  
##       chas              nox                rm               age         
##  Min.   :-0.2723   Min.   :-1.4644   Min.   :-3.8764   Min.   :-2.3331  
##  1st Qu.:-0.2723   1st Qu.:-0.9121   1st Qu.:-0.5681   1st Qu.:-0.8366  
##  Median :-0.2723   Median :-0.1441   Median :-0.1084   Median : 0.3171  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.:-0.2723   3rd Qu.: 0.5981   3rd Qu.: 0.4823   3rd Qu.: 0.9059  
##  Max.   : 3.6648   Max.   : 2.7296   Max.   : 3.5515   Max.   : 1.1164  
##       dis               rad               tax             ptratio       
##  Min.   :-1.2658   Min.   :-0.9819   Min.   :-1.3127   Min.   :-2.7047  
##  1st Qu.:-0.8049   1st Qu.:-0.6373   1st Qu.:-0.7668   1st Qu.:-0.4876  
##  Median :-0.2790   Median :-0.5225   Median :-0.4642   Median : 0.2746  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.6617   3rd Qu.: 1.6596   3rd Qu.: 1.5294   3rd Qu.: 0.8058  
##  Max.   : 3.9566   Max.   : 1.6596   Max.   : 1.7964   Max.   : 1.6372  
##      black             lstat              medv        
##  Min.   :-3.9033   Min.   :-1.5296   Min.   :-1.9063  
##  1st Qu.: 0.2049   1st Qu.:-0.7986   1st Qu.:-0.5989  
##  Median : 0.3808   Median :-0.1811   Median :-0.1449  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.4332   3rd Qu.: 0.6024   3rd Qu.: 0.2683  
##  Max.   : 0.4406   Max.   : 3.5453   Max.   : 2.9865

The variables are more similar in scale and weight, which makes them easier to compare and estimate. They also all have mean zero.

Create a categorical variable of the crime rate in the Boston dataset (from the scaled crime rate). This variable shows the quantiles of the scaled crime rate and is now used instead of the previous continuous one.

# class of the boston_scaled object
class(boston_scaled)
## [1] "matrix"
# change the object to data frame
boston_scaled <- as.data.frame(boston_scaled)

# summary of the scaled crime rate
summary(boston_scaled$crim)
##      Min.   1st Qu.    Median      Mean   3rd Qu.      Max. 
## -0.419367 -0.410563 -0.390280  0.000000  0.007389  9.924110
# create a quantile vector of crim and print it
bins <- quantile(boston_scaled$crim)
bins
##           0%          25%          50%          75%         100% 
## -0.419366929 -0.410563278 -0.390280295  0.007389247  9.924109610
# create a categorical variable 'crime'
crime <- cut(boston_scaled$crim, breaks = bins, include.lowest = TRUE, label = c("low", "med_low", "med_high", "high"))

# look at the table of the new factor crime
table(crime)
## crime
##      low  med_low med_high     high 
##      127      126      126      127

Let’s drop the old crime rate variable from the dataset and replace it with the new categorical variable for crime rates - for clarity:

# remove original crim from the dataset
boston_scaled <- dplyr::select(boston_scaled, -crim)

# add the new categorical value to scaled data
boston_scaled <- data.frame(boston_scaled, crime)

Finally, the last step. 80 % of the data will become the training (train) set and the 20 % the test set. The actual predictions of new data are done with the test set.

# number of rows in the Boston dataset 
n <- nrow(boston_scaled)

# choose randomly 80% of the rows
ind <- sample(n,  size = n * 0.8)

# create train set
train <- boston_scaled[ind,]

# create test set 
test <- boston_scaled[-ind,]

3. Linear discriminant analysis

Tasks 5 and 6

Now let’s fit the linear discriminant analysis on the train set. LDA is a generalization of Fisher’s linear discriminant, a method used in statistics, pattern recognition and machine learning to find a linear combination of features that characterizes or separates two or more classes of objects or events (as explained by everyone’s fav source).

We will use the categorical crime rate as the target variable and all the other variables in the dataset as predictor variables.

# linear discriminant analysis
lda.fit <- lda(crime ~., data = train)

# print the lda.fit object
lda.fit
## Call:
## lda(crime ~ ., data = train)
## 
## Prior probabilities of groups:
##       low   med_low  med_high      high 
## 0.2376238 0.2574257 0.2450495 0.2599010 
## 
## Group means:
##                   zn      indus         chas        nox          rm
## low       0.95629007 -0.8807867 -0.149294685 -0.8786138  0.41241787
## med_low  -0.08629773 -0.2941301 -0.007331936 -0.5770663 -0.09918927
## med_high -0.39022525  0.2106069  0.244663893  0.3629907  0.14639632
## high     -0.48724019  1.0149946 -0.047351911  0.9974409 -0.34817594
##                 age        dis        rad        tax     ptratio
## low      -0.8575282  0.8600479 -0.6887728 -0.7235527 -0.43463000
## med_low  -0.3061982  0.3659060 -0.5589261 -0.5096837 -0.06828869
## med_high  0.3679876 -0.3792863 -0.3925568 -0.2940621 -0.36344874
## high      0.7956798 -0.8418342  1.6596029  1.5294129  0.80577843
##               black       lstat       medv
## low       0.3899604 -0.77425474  0.4661234
## med_low   0.3247593 -0.15236716  0.0283201
## med_high  0.1270487 -0.03089723  0.2602402
## high     -0.8675639  0.84738417 -0.6875297
## 
## Coefficients of linear discriminants:
##                 LD1         LD2           LD3
## zn       0.12559524  0.82760902 -0.9555427636
## indus    0.02460583 -0.19270720  0.1961038012
## chas    -0.06567985 -0.08517354  0.1050450652
## nox      0.30354912 -0.84657169 -1.3434421787
## rm      -0.12382386 -0.05470384 -0.1510458391
## age      0.22966187 -0.12321029 -0.0004656232
## dis     -0.14135890 -0.31612002  0.2834796381
## rad      3.44385977  0.95424676 -0.0977116117
## tax      0.02082055 -0.18769321  0.5257460286
## ptratio  0.14882797  0.10555894 -0.1512685037
## black   -0.16989942 -0.01032241  0.0828311617
## lstat    0.18738606 -0.31471559  0.3812395597
## medv     0.20710278 -0.50258861 -0.1982959010
## 
## Proportion of trace:
##    LD1    LD2    LD3 
## 0.9568 0.0329 0.0103

The LDA calculates the probability of a new observation being classified as belonging to each class on the basis of the trained model, and assigns every observation to the most probable class.

# the function for lda biplot arrows
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "red", tex = 0.75, choices = c(1,2)){
  heads <- coef(x)
  arrows(x0 = 0, y0 = 0, 
         x1 = myscale * heads[,choices[1]], 
         y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
  text(myscale * heads[,choices], labels = row.names(heads), 
       cex = tex, col=color, pos=3)
}

# target classes as numeric
classes <- as.numeric(train$crime)

# plot the lda results
plot(lda.fit, dimen = 2, col = classes, pch = classes)
lda.arrows(lda.fit, myscale = 4)

Biplot is a visualisation chart that allows that allows us to clearly see some of the most outstanding or clear predictor vairables. It is clearly visible that accessibility to radial highways - rad - is the variable that is the most telling.

In order to assess the performance of the model in predicting the crime rate, let’s save the crime categories from the test set and then remove the categorical crime variable from the test dataset…

# save the correct classes from test data
correct_classes <- test$crime

# remove the crime variable from test data
test <- dplyr::select(test, -crime)

…and then predict the classes with the LDA model on the test data with the predict() function, and cross tabulate the results with the crime categories from the test set:

# predict classes with test data
lda.pred <- predict(lda.fit, newdata = test)

# cross tabulate the results
table(correct = correct_classes, predicted = lda.pred$class)
##           predicted
## correct    low med_low med_high high
##   low       17      13        1    0
##   med_low    3      16        3    0
##   med_high   1      12       13    1
##   high       0       0        1   21

The corss tabulation of the results tells us that the model predicts crime rate in the suburbs correctly (which is to be expected, since it was such a telling feature previously); the model has some problems in separating med_low from low, but overall it performs really well.

4. K-means clustering

Task 7

It’s time for data clustering. Let’s reload the Boston dataset and standardize it.

# center and standardize variables
boston_scaled <- scale(Boston)

# change the object to data frame
boston_scaled <- as.data.frame(boston_scaled)

The next step is to calculate the (Euclidean) distances between the observations, and to do that we’ll use a Euclidean distance matrix:

# euclidean distance matrix
dist_eu <- dist(Boston)

# look at the summary of the distances
summary(dist_eu)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   1.119  85.624 170.539 226.315 371.950 626.047

Now let’s perform the K-means clustering with K=3 and have a look at the plot (the last 5 columns):

# k-means clustering
km <-kmeans(Boston, centers = 3)

# plot the Boston dataset with clusters
pairs(Boston[6:10], col = km$cluster)

But is it optimal? How do we know what the optimal amount of clusters is?

Let’s take the within cluster sum of squares (WCSS) and look at the changes in it depending on the number of clusters. The optimal number of clusters shows as a sharp drop in total WCSS.

set.seed(123)

# determine the number of clusters
k_max <- 10

# calculate the total within sum of squares
twcss <- sapply(1:k_max, function(k){kmeans(Boston, k)$tot.withinss})

# visualize the results
qplot(x = 1:k_max, y = twcss, geom = 'line')

The optimal number of cluster seems to be 2, so let’s use that:

# k-means clustering
km <-kmeans(Boston, centers = 2)

# plot the Boston dataset with clusters
pairs(Boston[6:10], col = km$cluster)

We can also have a look at other columns:

pairs(Boston[7:14], col = km$cluster)

Again it looks like the same variables as before are the most distinctive: access to highways and property tax.

5. Bonus

Actually the super-bonus exercise, because it’s worth more points.

Run the code below for the (scaled) train data that you used to fit the LDA. The code creates a matrix product, which is a projection of the data points.

model_predictors <- dplyr::select(train, -crime)

# check the dimensions
dim(model_predictors)
## [1] 404  13
dim(lda.fit$scaling)
## [1] 13  3
# matrix multiplication
matrix_product <- as.matrix(model_predictors) %*% lda.fit$scaling
matrix_product <- as.data.frame(matrix_product)

Next, install and access the plotly package. Create a 3D plot (Cool!) of the columns of the matrix product by typing the code below.

# access the needed libraries:
library(plotly)
## 
## Attaching package: 'plotly'
## The following object is masked from 'package:MASS':
## 
##     select
## The following object is masked from 'package:ggplot2':
## 
##     last_plot
## The following object is masked from 'package:stats':
## 
##     filter
## The following object is masked from 'package:graphics':
## 
##     layout
plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers')

Adjust the code: add argument color as a argument in the plot_ly() function. Set the color to be the crime classes of the train set.

plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers', color = train$crime)

Draw another 3D plot where the color is defined by the clusters of the k-means.

plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers', color = km$centers)

Hmm. This is difficult to interpret?